Problem: A particle moves along the $x$ -axis. The function $x(t)$ gives the particle's position at any time $t\geq 0$ : $x(t)=t^4-2t^2-4$ What is the particle's velocity $v(t)$ at $t=1$ ? $v(1)=$
Answer: We have a function for the particle's position, and we need to find the particle's velocity. Since velocity is the rate of change of position, we need to find the derivative of $x(t)$. In other words, if $v(t)$ gives the particle's velocity at any time $t\geq 0$, then $v(t)=x'(t)$. Let's differentiate $x(t)$ to find $v(t)$ : $\begin{aligned} v(t)&=x'(t) \\\\ &=\dfrac{d}{dt}[t^4-2t^2-4] \\\\ &=4t^3-4t \end{aligned}$ To find the particle's velocity at $t=1$, we need to evaluate $v(1)$. $\begin{aligned} v({1})&=4({1})^3-4({1}) \\\\ &=0 \end{aligned}$ Since the velocity is zero, we know the particle is neither moving to the right nor to the left. The particle's velocity at $t=1$ is $0$. At $t=1$, the particle is neither moving to the right nor to the left.